Irodov 3.219: Every Element Pushes Along the Axis – Amit & Deepali Sinha Foundation

Problem Irodov 3.219

A thin circular wire loop of radius $R = 100$ mm carries a steady current $I = 1.00$ A. Find the magnetic induction

at the center of the loop;
at the point on the loop’s axis a distance $x = 100$ mm from the center.

The loop of radius R carrying I (sense chosen), its axis, and the axis point P a distance x from the center. — The loop of radius $R$ carrying $I$ (sense chosen), its axis, and the axis point $P$ a distance $x$ from the center.

Given

$I = 1.00$ A
$R = 100$ mm
$x = 100$ mm
$\mu_0 = 4\pi\times10^{-7}$ H/m

Find

B

at the center and at

P

Solution

Step 1The field of one element

The Biot-Savart law gives the field at a point a displacement $\vect r$ from a current element $I\,d\vect l$ .

d\vect B = \frac{\mu_0 I}{4\pi}\,\frac{d\vect l\times\vect r}{r^3} \tag{1}

Every element of a circular loop is perpendicular to the line joining it to any point of the axis, so the magnitude of (1) is simply

dB = \frac{\mu_0 I}{4\pi}\,\frac{dl}{r^2} \tag{2}

Step 2At the center: all elements add

At the center every element is a distance $r = R$ away and every $d\vect B$ points the same way along the axis, so the magnitudes (2) add over the whole circumference $2\pi R$ :

B_0 = \frac{\mu_0 I}{4\pi R^2}\oint dl = \frac{\mu_0 I}{4\pi R^2}\cdot 2\pi R

Numerically,

B_0 = \frac{4\pi\times10^{-7}\times1.00}{2\times0.100}\ \text{T}

B_0 = \frac{\mu_0 I}{2R} = 6.28\ \mu\text{T}

(3)

$\vect B_0$ lies along the axis, in the sense the right-hand rule gives from the current: fingers along $I$ , thumb along $\vect B_0$ .

Step 3On the axis: only the axial parts survive

The element d l at the top of the loop, its r to P, and d B normal to r, split into an axial part and a transverse part; the opposite element gives d B', whose transverse part cancels. — The element $d\vect l$ at the top of the loop, its $\vect r$ to $P$ , and $d\vect B$ normal to $\vect r$ , split into an axial part and a transverse part; the opposite element gives $d\vect B'$ , whose transverse part cancels.

At $P$ every element is the same distance away,

r = (R^2 + x^2)^{1/2}

and $d\vect B$ is perpendicular to $\vect r$ , tilted off the axis by the angle whose sine is $x/r$ . The diametrically opposite element gives the mirror image, so the transverse parts cancel in pairs and only the axial component survives, a fraction $R/r$ of (2):

dB_{\text{axial}} = \frac{\mu_0 I}{4\pi}\,\frac{dl}{r^2}\cdot\frac{R}{r}

All of these point the same way, so they add over the circumference:

B = \frac{\mu_0 I R}{4\pi r^3}\oint dl = \frac{\mu_0 I R}{4\pi r^3}\cdot 2\pi R

B = \frac{\mu_0 R^2 I}{2\,(R^2 + x^2)^{3/2}}

(4)

Step 4The number at

x = R

Here $x = R$ , so $(R^2 + x^2)^{3/2} = 2^{3/2}R^3$ and (4) is the center value divided by $2^{3/2}$ :

B = \frac{\mu_0 I}{2R}\cdot\frac{1}{2^{3/2}}

Numerically, with the center value (3),

B = \frac{6.283\ \mu\text{T}}{2.828}

B = \frac{\mu_0 I}{2^{5/2}R} = 2.22\ \mu\text{T}

(5)

It points along the axis in the same sense as $\vect B_0$ .

Check

At $x = 0$ , (4) gives $\mu_0 R^2 I/(2R^3) = \mu_0 I/(2R)$ , which is (3). Far away, $x \gg R$ , it falls as $\mu_0 R^2 I/(2x^3)$ , the inverse cube of a dipole, and the units of $\mu_0 I/R$ are T $\cdot$ m/A times A/m, which is tesla.

Remark

With the loop’s magnetic moment written as $p_m = I\pi R^2$ , the far field on the axis becomes $\mu_0 p_m/(2\pi x^3)$ . It has the same form as the axial field $p/(2\pi\varepsilon_0 x^3)$ of an electric dipole with $1/\varepsilon_0$ replaced by $\mu_0$ . Already at $x = R$ the field has dropped to $1/2^{3/2} \approx 0.35$ of its central value.